Solution Manual For Chemistry: Principles and Reactions, 8th Edition

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Contents
Preface v
Lecture Schedule vii
Chapter 1 Matter and Measurements 1
Chapter 2 Atoms, Molecules, and Ions 11
Chapter 3 Mass Relations in Chemistry; Stoichiometry 21
Chapter 4 Reactions in Aqueous Solution 37
Chapter 5 Gases 49
Chapter 6 Electronic Structure and the Periodic Table 61
Chapter 7 Covalent Bonding 71
Chapter 8 Thermochemistry 83
Chapter 9 Liquids and Solids 95
Chapter 10 Solutions 105
Chapter 11 Rate of Reaction 119
Chapter 12 Gaseous Chemical Equilibrium 133
Chapter 13 Acids and Bases 147
Chapter 14 Equilibria in Acid-Base Solutions 161
Chapter 15 Complex Ions and Precipitation Equilibra 177
Chapter 16 Spontaneity of Reaction 189
Chapter 17 Electrochemistry 201
Chapter 18 Nuclear Reactions 219
Chapter 19 Complex Ions 229
Chapter 20 Chemistry of the Metals 237
Chapter 21 Chemistry of the Nonmetals 245
Chapter 22 Organic Chemistrry 255
Chapter 23 Organic Polymers: Natural and Synthetic 265
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Preface
This manual starts off with a section entitled “Lecture Schedule,” which you may find helpful in adapting the
text to your class schedule. Beyond that, the manual is organized by text chapters. For each chapter, we
include three different features:
1. “Lecture Notes,” which suggest the amount of time that we devote to each chapter and the topics we
emphasize. Included are detailed lecture outlines (from our own lectures) that may serve as a guide
for your lectures. At a minimum, they indicate how we cover topics and how successive topics can be
integrated.
2. A list of demonstrations illustrating topics in the chapter. These are taken from three sources:
• The manual Tested Demonstrations in Chemistry (1994), Volumes I and II, compiled and edited by
by George Gilbert by arrangement with the Journal of Chemical Education. These are coded as
“GILB” with the experiment number (e.g., M 12)
• Chemical Demonstrations (1983-1992), Volumes 1-4, published by Bassam Shakashiri with many
collaborators and contributors. These are listed as “SHAK” followed by the volume and page
reference.
• Demonstrations described in the Journal of Chemical Education with the journal reference.
3. Answers and detailed solutions to:
• The summary problem at the end of each chapter
• Odd numbered text problems
• Challenge problems at the end of each problem set
Note that Appendix 6 has answers to all the even-numbered problems. Detailed solutions to many of
these problems are in the Student Solutions Manual available from Cengage Learning, Brooks/Cole.
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Lecture Schedule
Unlike other general chemistry texts, ours can be covered in its entirety in a one-year course. A reasonable
schedule appears below. Further comments on the time you should devote to each chapter are in the body
of this manual. The underlying assumption is that you are teaching 14-week semesters with two 50-minute
lectures per week. If three class periods are devoted to examinations each semester, that leaves 25 for
covering material. On that basis, you are able to complete Chapter 10 (Solutions) in the first semester. The
seond semester will then start with Chapter 11 (Rate of Reaction).
FIRST SEMESTER SCHEDULE
Week Lecture Topic
1 1 Chapter 1 (Matter and Measurements)
2 Chapter 1
2 3 Chapter 2 (Atoms, Molecules, and Ions)
4 Chapter 2
3 5 Chapter 3 (Mass Relations in Chemistry; Stoichiometry)
6 Chapter 3
4 7 Chapter 3
8 EXAM I
5 9 Chapter 4 (Reactions in Aqueous Solution)
10 Chapter 4
6 11 Chapter 4
12 Chapter 5 (Gases)
7 13 Chapter 5
14 Chapter 6 (Electronic Structure and the Periodic Table)
8 15 Chapter 6
16 Chapter 6
9 17 EXAM II
18 Chapter 7 (Covalent Bonding)
10 19 Chapter 7
20 Chapter 7
11 21 Chapter 8 (Themochemistry)
22 Chapter 8
12 23 Chapter 9 (Liquids and Solids)
24 Chapter 9
13 25 Chapter 9
26 EXAM III
14 27 Chapter 10 (Solutions)
28 Chapter 10
vii

viii Lecture Schedule
SECOND SEMESTER SCHEDULE
Week Lecture Topic
1 1 Chapter 11 (Rate of Reaction)
2 Chapter 11
2 3 Chapter 11
4 Chapter 12 (Gaseous Chemical Equilibrium)
3 5 Chapter 12
6 Chapter 13 (Acids and Bases)
4 7 Chapter 13
8 Chapter 13
5 9 EXAM I
10 Chapter 14 (Equilibria in Acid-Base Solutions)
6 11 Chapter 14
12 Chapter 15 (Complex Ions)
7 13 Chapter 15
14 Chapter 16 (Precipitation Equilibria)
8 15 Chapter 17 (Spontaneity of Reaction)
16 Chapter 17
9 17 EXAM II
18 Chapter 18 (Electrochemistry)
10 19 Chapter 18
20 Chapter 18
11 21 Chapter 19 (Nuclear Chemistry)
22 Chapter 20 (Chemistry of the Metals)
12 23 Chapter 20
24 EXAM III
13 25 Chapter 21 (Chemistry of the Nonmetals)
26 Chapter 21
14 27 Chapter 22 (Organic Chemistry)
28 Chapter 23 (Organic Polymers: Natural and Synthetic)

Lecture Schedule ix
If you want to use lecture time for review, for going over assigned problems, or for doing a large number of
demonstrations, you will have trouble keeping up with this schedule. As you’ve almost certainly learned by
now, the solution to this problem is not to talk faster. Judicious deletions work better. It’s been said, and
wisely, that the secret of giving a good lecture is knowing what to leave out. Possible candidates include:
• Introductory material on matter in Chapter 1 and atomic theory in Chapter 2. The chances are your
students have been exposed to this material more than once in high school and understood it reasonably
well the first time.
• Boyle’s and Charles’s laws in Chapter 5. We start the chapter by writing the ideal gas law and go on
from there.
• The First Law discussion in Chapter 8. Quite frankly, this has very little to do with chemistry. Students
will not be irreparably damaged if they are unaware of the distinction between H and E.
• The discussion of colligative properties in Chapter 10 could be shortened. Raoult’s law could easily be
omitted.
• Reaction mechanisms in Chapter 11. Students have a lot of trouble with this. We are not sure it is worth
the effort.
• Polyprotic acids in Chapter 13.
• The Second Law discussion in Chapter 17.
Beyond these selective omissions, some instructors may want to delete one or another of the descriptive
chapters at the end of the text (Chapters 20–22). If, in that way, you can squeeze out a couple of lectures,
they can well be spent on Chapter 12 (three lectures instead of two) and Chapter 19 (two lectures instead
of one).
Textbook authors sometimes tell you that chapters can be covered in almost any order, depending on
your preference. This isn’t really true for this textbook, or any other with structural integrity. It can be done,
but only with very careful additions and deletions of material. Suppose, for example, you want to cover
Precipitation Equilibria (Chapter 16) immediately after Acid-Base Equilibria (Chapter 14). Keep in mind
that an understanding of formation constants (Complex Ions, Chapter 15) is assumed when methods of
dissolving precipitates are considered in Section 16.2 of Chapter 16.

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MATTER AND
MEASUREMENTS
LECTURE NOTES
This material ordinarily requires two lectures (100 minutes), allowing for a 10–15 minute introduction to the
course in the first lecture. If you’re in a hurry, this can be cut to 1 1
2 lectures by discussing only quantitative
material (significant figures, unit conversions, density, solubility).
A few points to keep in mind:
• Virtually all of your students will be familiar with the metric system and prefixes. It may be worth
discussing the rationale for SI, but you don’t have to dwell on it.
• Students readily learn the rules of significant figures, but typically ignore them after Chapter 1. It may
help to emphasize that these are common-sense (albeit, approximate) rules for estimating experimental
error.
• Many (typically, the weaker) students resist using conversion factors, preferring instead a rote method.
It may be useful to point out that conversion factors will be a recurring tool throughout the text, so are
well worth learning at this point.
• Students often have trouble with solubility calculations. The approach in the text involves conversions
(Example 1.8). The solubility is considered to be a conversion factor relating grams of solute to grams
of solvent.
Lecture 1
I. Types of Substances
A. Elements
Cannot be broken down into simpler substances. Examples: nitrogen, lead, sodium, arsenic.
Symbols: N, Pb, Na, As.
B. Compounds
Contain two or more elements with fixed mass percents. Glucose: 40.00% C, 6.71% H, 53.29%
O. Sodium chloride: 39.34% Na, 60.66% Cl.
C. Mixtures
Homogeneous (solutions) vs. heterogeneous. Separation by filtration, distillation.
II. Measured Quantities
A. Length
Base unit is the meter. 1 km = 103 m; 1 cm = 102 m; 1 mm = 103 m; 1 nm = 109 m. Dimensions
of very tiny particles will be expressed in nanometers.
1

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2 Chapter 1
B. Volume
1 L = 103 mL = 103 cm3 = 103 m3 . Buret, pipet, volumetric flask.
C. Mass
1 kg = 103 g; 1 mg = 103 g. Two different kinds of balances will be used in the lab. An analytical
balance (±0.001 g) should be used only for accurate, quantitative work.
D. Temperature
tF = 1.8 tC + 32 ; TK = tC + 273.15.
Convert 68F to C and K: tC = (68 32)/1.8 = 20C TK = 293
Lecture 2
III. Experimental Error; Significant Figures
Suppose an object is weighed on a crude balance to ±0.1 g and the mass is found to be 23.6 g.
This quantity contains three significant figures, that is, three experimentally significant digits. With an
analytical balance, the mass might be 23.582 g (five significant figures).
A. Counting significant figures
1. Volume of liquid = 24.0 mL; three significant figures. Zeroes at the end of the measured
quantity are significant when they follow nonzero digits.
2. Volume = 0.0240 L; three significant figures (note that 0.0240 L = 24.0 mL). Zeroes at the
beginning of a measured quantity are not significant when they precede nonzero digits.
B. Multiplication and division
Keep only as many significant figures as there are in the least precise quantity. Density of a piece
of metal weighing 36.123 g with a volume of 13.4 mL = ?
density = 36.123 g
13.4 mL = 2.70 g/mL
C. Addition and subtraction
Keep only as many digits after the decimal point as there are in the least precise quantity. Add
1.223 g of sugar to 154.5 g of coffee:
total mass = 1.2 g + 154.5 g = 155.7 g
Note that the rule for addition and subtraction does not apply to significant figures. The number of
significant figures may well decrease after subtraction.
mass beaker + sample = 52.169 g (five significant figures)
mass empty beaker = 52.120 g (five significant figures)
mass sample = 0.049 g (two significant figures)
D. Exact numbers
One liter means 1.000000... L.

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Matter and Measurements 3
IV. Conversion Factors
A. One-step conversion
A rainbow trout is measured to be 16.2 inches long. What is its length in centimeters?
length in cm = 16.2 in ⇥ 2.54 cm
1 in = 41.1 cm
Note the cancellation of units. To convert from centimeters to inches, use the conversion factor
1 in = 2.54 cm. (Here, there are exactly 2.54 cm in one inch.)
B. Multiple conversion factors
A thrown baseball has speed 89.6 miles per hour. What is its speed in meters per second?
1 mile = 1.609 km = 1.609 ⇥ 103 m; 1 h = 3600 s
speed =
✓
89.6 mile
h
◆
⇥
⇣
1.609 ⇥ 103 m
mile
⌘
⇥
✓ 1 h
3600 s
◆
= 40.0 m/s
V. Properties of Substances
Distinguish between intensive and extensive, and between chemical and physical.
A. Density
An empty flask weighs 22.138 g. Pipet 5.00 mL of octane into the flask, producing a total mass of
25.598 g. What volume is occupied by ten grams of octane?
d = 3.460 g/5.00 mL = 0.692 g/mL
V = 10.00 g ⇥ 1 mL
0.692 g = 14.5 mL
Note that for density calculations, 1 mL = 1 cm3 .
B. Solubility
This is often expressed as grams of solute per 100 g of solvent.
Solubility of sugar at 20C = 210 g sugar/100 g water.
A solution containing 210 g sugar/100 g water is saturated.
A solution containing less than 210 g sugar/100 g water is unsaturated.
A solution containing more than 210 g sugar/100 g water is supersaturated.
1. How much water is required to dissolve 52 g of sugar at 20C?
52 g sugar ⇥ 100 g water
210 g sugar = 25 g water

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4 Chapter 1
2. A solution at 20C contains 25 g sugar and 125 g water. Is it unsaturated, saturated or super-
saturated?
mass sugar/100 g water = 25 g sugar
125 g water ⇥ 100 g water = 20 g sugar (unsaturated)
DEMONSTRATIONS
1. Scientific method: GILB H 29
2. Decomposition of mercury(II) oxide: GILB A 8
3. Separation of a mixture: GILB A 14
4. Reaction of sodium with chlorine: GILB A 24, A 25; SHAK 1 61; J. Chem. Educ. 73 539 (1996)
5. Chromatography: GILB Q 3, Q 13
6. Significant figures: J. Chem. Educ. 69 497 (1992)
7. Density of liquids: GILB C 13; SHAK 3 229
8. Supersaturation: GILB F 11; SHAK 1 27
SUMMARY PROBLEM
(a) K, Mn, O
(b) density, melting point, solubility, color
(c) mass = 2.703 g
cm3 ⇥ 48.7cm3 = 132 g
(d) 2.703 g
cm3 ⇥ 1 lb
454 g ⇥ (2.54)3 cm3
13 in3 ⇥ (12)3 in3
1 ft = 169 lb/ft3
(e) F = 9
5 (C) + 32 = 9
5 (2.40 ⇥ 102 ) + 32 = 464F
(2.40 ⇥ 102 ) + 273 = 513 K
(f) 38.5 g H2 O ⇥ 6.38 g KMnO4
100 g H2 O = 2.46 g KMnO4
(g) At 60C: 65.0 g H2 O ⇥ 25 g KMnO4
100 g H2 O = 16 g KMnO4
The solution is unsaturated.
At 20C: 65.0 g H2 O ⇥ 6.38 g KMnO4
100 g H2 O = 4.15 g KMnO4 can be dissolved.
The solution is supersaturated.

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Matter and Measurements 5
(h) 55.0 g H2 O ⇥ 25 g KMnO4
100 g H2 O = 14 g KMnO4 can be dissolved at 60C.
Yes, all the KMnO4 added will dissolve.
55.0 g H2 O ⇥ 6.38 g KMnO4
100 g H2 O = 3.51 g KMnO4 can be dissolved at 20C.
No, not all the KMnO4 will dissolve.
10.0 g - 3.51 g = 6.5 g KMnO4 will remain undissolved.
PROBLEMS
1. (a) mixture (b) element (c) mixture (d) compound
3. (a) solution (b) solution (c) heterogeneous mixture
5. (a) distillation (b) filtration (c) gas chromatography
7. (a) Ti (b) P (c) K (d) Mg
9. (a) mercury (b) silicon (c) sodium (d) iodine
11. (a) balance (b) thermometer (c) graduated cylinder
13. tF = 1.8(52) + 32 = 126F; tK = 52 + 273.15 = 325 K
15. tC = (85.0 32) ⇥ 5
9 = 29.4C ; solid
17. (a) 3 (b) ambiguous (c) 4
(d) exact (e) 5
19. (a) 7.49 g (b) 298.69 cm (c) 1 ⇥ 101 lb (d) 12.0 oz
21. (a) 1.325 ⇥ 102 cm (b) 8.83 ⇥ 104 km (c) 6.432 ⇥ 109 nm
23. (c)
25. 10,000: ambiguous 1.71 ⇥ 105 ft2 : 3 $22.00: exact 20%: ambiguous
27. (a) 80.0 (b) 0.7615 (c) 14.712
(d) 0.03 (e) 1.5 ⇥1022

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6 Chapter 1
29. 4⇡(4.30 cm)3
3 = 333 cm3 ; 4⇡(4.33 cm)3
3 = 3.40 ⇥ 102 cm3 ; 7 cm3
31. (a) 303 m = 0.303 km < 303 ⇥ 103 km (b) 500 g = 0.500 kg
(c) 1.50 cm3 = 1.50 ⇥ 1021 nm3 > 1.50 ⇥ 103 nm3
33. (a) 22.3 mL ⇥ 1 L
103 mL = 2.23 ⇥ 102 L (b) 22.3 cm3 ⇥ 1 in3
(2.54 cm)3 = 1.36 in3
(c) 22.3 mL ⇥ 1 L
103 mL ⇥ 1.057 qt
1 L = 0.0236 qt
35. (a) 19.2 hands ⇥
1
3 ft
1 hand = 6.40 ft
(b) 17.8 hands ⇥
1
3 ft
1 hand ⇥ 12 in
1 ft ⇥ 1 m
39.37 in = 1.81 m
(c) 20.5 hands ⇥
1
3 ft
1 hand + 3.0 ft = 9.8 ft
37. 2.0 acre ⇥ 4.356 ⇥ 104 ft2
1 acre ⇥ (12)2 in2
1 ft2 ⇥ 1 m2
(39.37)2 in2 ⇥ 1 hectare
104 m2 = 0.81 hectare
39. 5.0 mi
1 mile ⇥ 0.25 min
1 Eng lap = 1.2 min
0.50 km ⇥ 1 mi
1.609 km ⇥ 5.0 min
1 mi = 1.6 min
41. (a) 3.0 qt plasma ⇥ 1 L
1.057 qt ⇥ 1000 mL
1 L ⇥ 0.080 mL alcohol
100 mL plasma = 2.3 mL
(b) 3.0 qt plasma ⇥ 1 L
1.057 qt ⇥ 1000 mL
1 L ⇥ 0.10 mL alcohol
100 mL plasma = 2.8 mL
(c) 2.8 mL 2.3 mL = 0.5 mL
43. 235 kJ
250 mL ⇥ 103 J
1 kJ ⇥ 1 cal
4.18 J ⇥ 1 kcal
103 cal ⇥ 1000 mL
1.057 qt ⇥ 1 qt
4 cups = 53.2 kcal/cup
45. 252 g
0.750 ⇥ 225 mL = 1.49 g/mL

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Matter and Measurements 7
47. Vmethanol = 43.7 g ⇥ 1 mL
0.791 g = 55.2 mL , Vslug = 59.7 mL 55.2 mL = 4.5 mL
Therefore, dslug = 25.17 g
4.5 mL = 5.6 g/mL
49. (8.0 ⇥ 7.0 ⇥ 0.75) ft3 ⇥ (12)3 in3
1 ft3 ⇥ (2.54)3 cm3
1 in3 ⇥ 1.00 g
1 cm3 ⇥ 1 kg
1000 g = 1.2 ⇥ 103 kg
51. Volume of air = volume of room
V = 55 kg O2 ⇥ 1000 g
1 kg ⇥ 1 L
1.31 g ⇥ 100 L
21 L = 2.0 ⇥ 105 L
53. At 30C: maximum amount of MgSO4 that can be dissolved = 25.0 g H2 O ⇥ 38.9 g MgSO4
100 g H2 O = 9.72 g
The solution is unsaturated.
(9.50 + 1.00 g) - 9.72 = 0.78 g will precipitate out
55. (a) 46 g H2 O ⇥ 16 g NaHCO3
100 g H2 O = 7.4 g NaHCO3 can be dissolved
9.2 g in the mixture > 7.4 g, thus the solution is not homogeneous.
9.2 g – 7.4 g = 1.8 g NaHCO3 are undissolved.
(b) 9.2 g NaHCO3 ⇥ 100 g H2 O
9.6 g NaHCO3
= 96 g H2 O needed to dissolve
96 g 46 g = 5.0 ⇥ 101 g H2 O needs to be added.
57. 57.0 g 25.0 g = 32.0 g of Pb(NO3 )2 dissolves in 64.0 g H2 O at 10C. Solubility is
32.0 g Pb(NO3 )2
64.0 g H2 O = 1.00 g PbNO3 )2
2.00 g H2 O = 50.0 g Pb(NO3 )2
100.0 g H2 O
59. (a) physical (b) physical (c) physical (d) chemical
61. V = 35 ft ⇥ 43 ft ⇥ 28 in ⇥ 1 ft
12 in = 3.5 ⇥ 103 ft3
3.5 ⇥ 103 ft3 ⇥ (12 in)3
(1 ft)3 ⇥ (2.54 cm)3
(1 in)3 = 9.9 ⇥ 107 cm3
mass = 9.9 ⇥ 107 cm3 ⇥ (0.35 g)
(1 cm)3 ⇥ 1 lb
453.6 g = 7.7 ⇥ 104 lbs

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8 Chapter 1
63. 108 carats ⇥ 0.200 g
1 carat ⇥ 1 lb
454 g = 0.0476 lb
0.0476 lb ⇥ 1 cm3
3.51 g ⇥ 454 g
1 lb ⇥ 1 in3
(2.54)3 cm3 = 0.376 in3
65. 153.2 g ⇥ 1 cm3
4.55 g = 33.7 cm3 = V; 33.7 = ⇡r2 (7.75); r = 1.18 cm; d = 2.35 cm
67. (a) Chemical properties show the behavior of the species in a reaction; physical properties
are intrinsic qualities.
(b) Distillation vaporizes the liquid; filtration removes the solid.
(c) The solute is a component of the solution.
69. The bottom layer is Hg; the middle layer is Pb; the top layer is ethyl alcohol.
71. (a) ⇡ 115 g; supersaturated (b) ⇡ 30 g; unsaturated
(c) Dissolve 30 g of compound in 100 g H2 O.
73. (a) See Figure in the answers to the problems in Appendix 5.
(b) y
x = 100J 0J
78C (117.0C) = 0.512
(c) 60J
(d) J = 0.51(C) + 60
74. 31.5 gal ⇥ 4 qt
1 gal ⇥ 1 L
1.057 qt ⇥ 103 m3
1 L ⇥ 1 km3
109 m3 = 1.2 ⇥ 1010 km3
area = 1.2 ⇥ 1010 km3
100 nm ⇥ 1 ⇥ 1012 nm
1 km = 1.2 km2
75. V = 12.0 g ⇥ 1 cm3
2.70 g = ⇡(0.254 cm)2`; ` = 21.9 cm
76. 8.50 ⇥ 103 L
1 d ⇥ 1 m3
103 L ⇥ 7.0 ⇥ 106 g Pb
1 m3 ⇥ 0.75 ⇥ 0.50 ⇥ 365 d
1 yr = 8.1 ⇥ 103 g Pb

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Matter and Measurements 9
78. mass of Hg in cylinder B = 145.20 g
mass of Hg + metal in cylinder A = 92.60 g
mass of metal = 145.20 g 92.60 g = 52.60 g
volume of cylinder A = volume of cylinder B = volume of Hg in cylinder B
= 145.2 g ÷ 13.6 g/mL = 10.7 mL
mass of Hg in cylinder A = 92.60 g – 52.60 g = 40.0 g
volume of Hg in cylinder A = 40.0 g ÷ 13.6 g/mL = 2.94 mL
volume of metal = volume of cylinder A - volume of Hg in cylinder A = 10.7 mL 2.94 mL = 7.76 mL
density of metal = 52.60 g/7.76 mL = 6.78 g/mL

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ATOMS, MOLECULES
AND IONS
LECTURE NOTES
Students find this material relatively easy to assimilate; it’s almost entirely qualitative. On the other hand,
there’s a lot of memorizing (sorry, learning) to do. This chapter is coverable in two lectures.
Some general observations:
• Material in Sections 2.1–2.3 is generally well covered in high-school chemistry courses; no need to
dwell on it.
• Students need to know the molecular formulas of the elements (Figure 2.13), the charges of ions with
noble-gas structures and the names and formulas of the common polyatomic ions (Table 2.2). The
charges of transition-metal ions will be covered later, in Chapter 4.
• Naming compounds requires students to distinguish between ionic and molecular substances. It helps to
point out that binary molecular compounds are composed of two nonmetals. Almost all ionic compounds
contain a metal cation combined with a nonmetal anion or negatively charged polyatomic ion. The flow
charts shown in Figures 2.18 and 2.19 should help visual learners.
• The periodic table will be discussed in greater detail later in the text (Chapter 6).
Lecture 1
I. Atomic Theory
A. Elements
Postulates: Elements consist of tiny particles called atoms, which retain their identity in reactions.
In a compound, atoms of two or more elements combine in a fixed ratio of small whole numbers
(e.g., 1:1, 2:1, etc.).
B. Components
relative mass relative charge location
proton 1 +1 nucleus
neutron 1 0 nucleus
electron 0.0005 1 outside
C. Atomic number
It is the number of protons in the nucleus or the number of electrons in a neutral atom. This is
characteristic of a particular element: all H atoms have one proton, all He atoms have two protons,
etc.
11

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